Here's a parameterization of a cylinder, for $0 < \theta < 2\pi$ : $\vec{v}(\theta, z) = (r\cos(\theta), r\sin(\theta), z)$ What is the outward-pointing vector normal to the area element of this cylinder given $r = 5$, $\theta = \dfrac{3\pi}{2}$, and $z = 1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(0, -5, 0)$ (Choice B) B $(0, 5, 0)$ (Choice C) C $(-4, -3, 0)$ (Choice D) D $(4, 3, 0)$
Solution: The vector normal to the area element describes what's perpendicular to the surface, and the vector's magnitude represents the area of a tiny rectangle along the parameterization. After we compute it, we need to check whether it's pointing inwards or outwards. $\text{normal to area element} = \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial z}$ Let's calculate the cross product. $\begin{aligned} \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial z} &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ -r\sin(\theta) & r\cos(\theta) & 0 \\ \\ 0 & 0 & 1 \end{pmatrix} \\ \\ &= r\cos(\theta) \hat{\imath} + r\sin(\theta) \hat{\jmath} \end{aligned}$ Plugging in $r = 5$, $\theta = \dfrac{3\pi}{2}$, and $z = 1$, we get the vector normal to the area element: $\begin{aligned} \dfrac{\partial \vec{v}}{\partial \theta} \times \dfrac{\partial \vec{v}}{\partial z} &= \left( 5\cos\left(\dfrac{3\pi}{2}\right), 5\sin\left(\dfrac{3\pi}{2}\right), 0 \right) \\ \\ &= (0, -5, 0) \end{aligned}$ The final step is to check whether this points inwards or outwards. The cylinder defined by $\vec{v}$ has an outward facing normal vector that points directly away from the origin, but with a zero $z$ -component. Because $\theta = \dfrac{3\pi}{2}$ corresponds to a zero $x$, negative $y$, and zero $z$, we should have a zero $x$, negative $y$, and zero $z$. Our calculation matches the outward normal vector. Therefore, the outward-pointing vector normal to the area element is $(0, -5, 0)$.